The equation of a circle is given below. $(x+4)^{2}+(y-6)^{2} = 48$ What is its center? $($
Solution: Standard equation of the circle A circle is the collection of all points at a distance ${r}$ from a center $({h},{k})$. We can use the Pythagorean theorem to write an equation to relate the center and radius. $({h},{k})$ ${r}$ $x-{h}$ $y-{k}$ $(x, y)$ $\begin{aligned} a^2+b^2&=c^2\\\\ (x - {h})^2 + (y - {k})^2 &= {r}^2 \end{aligned}$ Rewriting the given equation We can rewrite the given equation as: $\begin{aligned}(x+4)^{2}+(y-6)^{2} &= 48\\\\ (x - {(-4)})^2 + (y - {6})^2 &= {48}\end{aligned}$ Finding the center According to the rewritten equation, we can see that the center of the circle is $({-4}, {6})$. Finding the radius According to the standard equation of the circle, we get that ${r^2}={48}$. Rounding the radius to two decimal places, we get that $r={\sqrt{48}}\approx {6.93}$. Summary The circle is centered at $(-4, 6)$. The circle has an approximate radius of $6.93$ units.